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2 years ago

A truck drives 80 km in 2 hours. The truck stops for 1 hr and continues driving for 100

Physics

1 answer:

bulgar [2K]2 years ago

7 0

Answer:

It Crashed Because it was speeding LOL

Explanation:

It was speeding so cops put spices and truck Crashed

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How are eras and periods of the geologic time scale named?

Natalka [10]

Answer:

mark me brainliest plz

Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

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2 years ago

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What happens when the dew point and the temperature are the same?

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The value of relative humidity becomes 100% leading to condensation of water vapor in the air into water droplets or water (dew)

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3 years ago

Monochromatic light from a point source is directed toward the sharp edge of a solid object. How does diffraction change the ima

Anna71 [15]

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3 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

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2 years ago

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m

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3 years ago

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